In LC circuit the inductance $\mathrm{L}=40 \mathrm{mH}$ and capacitance $\mathrm{C}=100 \mu \mathrm{~F}$. If a voltage $\mathrm{V}(\mathrm{t})= 10 \sin (314 \mathrm{t})$ is applied to the circuit, the current in the circuit is given as

(C) $0.52 \cos 314 \mathrm{t}$

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Electromagnetic Induction

In LC circuit the inductance

L = 40 mH

and capacitance

C = 100 μ F

. If a voltage

V (t) = 10 sin (314 t)

is applied to the circuit, the current in the circuit is given as

(A) $5.2 cos 314 t$
(B) $0.52 sin 314 t$
(C) $0.52 cos 314 t$ ✓ Correct
(D) $10 cos 314 t$

Answer: (C)

Solution

With no resistance the reactances are

X_{L} = ω L = 314 \times 0.04 \approx 12.6 Ω

and

X_{C} = \frac{1}{ω C} = \frac{1}{314 \times 100 \times 1 0 ^{- 6}} \approx 31.8 Ω

, so the net reactance is

X_{C} - X_{L} \approx 19.3 Ω

. Because

X_{C}

dominates, the current leads the applied voltage by

9 0^{\circ}

, and its amplitude is

\frac{V _{0}}{Z} = \frac{10}{19.3} \approx 0.52 A

. The current is therefore

0.52 cos (314 t)

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option

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